So what? The probability that Team A and Team B are in the top two is 1 10...(5 points).
(2) The possible values of the random variable X are 0, 1, 2, 3 (6 points).
P(X=0)= 2×4! 5! = 2 5 ; P(X= 1)= 3×2×3! 5! = 3 10 ; P(X=2)= 2×2! ×3×2! 5! = 1 5 ; P(X=3)= 2×3! 5! = 1 10
So the distribution list of x is
x 0 1 23 p253 10 1 5 1 10∴ex = 0×25+ 1×3 10+2× 1 5+3×65538。